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Homework Help: Tangent line of curve of intersection

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the parametric equations of the line tangent to the curve of intersection of the paraboloid
    z = x² + y² and the ellipsoid 4x² + y² + z² = 9 at the point ( -1, 1, 2 ).

    2. Relevant equations

    Probable use of the gradient vector (as this is the chapter we are in)

    3. The attempt at a solution

    I have found what I believe to be the curve of intersection by setting the equations of the surfaces equal to each other:

    x² + y² - z = 4x² + y² + z² - 9
    9 = 3x² + z² + z
    F(x,y,z) = 3x² + z² + z - 9 (as a function)

    And the corresponding gradient vector:
    < 6x, 0, 2z + 1 >

    But I cannot figure out how to convert this from cartesian to polar (or if I even need to) and get the correct answer. I've tried several paths (i.e. substituting x=rcosσ) but my answer never matches the expected answer, which contains no trigonometry. Does the correct approach involve substituting x² + y² for z? This seems to be the most logical approach but I haven't been able to apply it correctly.
  2. jcsd
  3. Mar 18, 2010 #2

    Filip Larsen

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    It is unclear to me why you want the tangent line in polar coordinates. The text seems only to call for a parametric line, eg. [itex]\bar{r} = \bar{k} t +\bar{r}_0[/itex].
  4. Mar 18, 2010 #3
    Your right! I was thinking of a vector equation. Then I suppose the parametric equations would look something like: x = -1 +3t², y = 1, z = 3 + t² ? But that still doesn't contain the expected y component, I must be doing something wrong yet.
  5. Mar 18, 2010 #4

    Filip Larsen

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    Gold Member

    A line parametric in, say, t should be linear in t, otherwise its not a line. Think about which point the line passes through and what vector direction it "follows" from that point as t varies.
  6. Mar 18, 2010 #5
    I remember why I wanted the vector equation. The only way I can think to find the tangent line is by taking the derivative of a vector equation.
  7. Mar 18, 2010 #6
    Never mind. By crossing the gradients I found the tangent vector. This makes perfect sense, it was so simple :D :D
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